. Find $\lim _ { y \rightarrow 0 } \frac { \sqrt { 15 + y } - \sqrt { 15 } } { y }$
A) $\frac { \sqrt { 15 } } { 30 }$
B) $\frac { \sqrt { 15 } } { 15 }$
C) 0
D) $\frac { \sqrt { 15 } } { 2 }$
E) limit does not exist

Question

Quizplus · Accepted Answer

To solve this limit, we can multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{15+y} + \sqrt{15}$, to eliminate the square root in the numerator. This gives:$$\lim _ { y \rightarrow 0 } \frac { \sqrt { 15 + y } - \sqrt { 15 } } { y } \cdot \frac { \sqrt { 15 + y } + \sqrt { 15 } } { \sqrt { 15 + y } + \sqrt { 15 } }$$Simplifying the numerator using the difference of squares formula, we get:$$\lim _ { y \rightarrow 0 } \frac { 15 + y - 15 } { y ( \sqrt { 15 + y } + \sqrt { 15 } ) } = \lim _ { y \rightarrow 0 } \frac { y } { y ( \sqrt { 15 + y } + \sqrt { 15 } ) }$$Canceling $y$ from the numerator and denominator:$$\lim _ { y \rightarrow 0 } \frac { 1 } { \sqrt { 15 + y } + \sqrt { 15 } }$$Substituting $y = 0$:$$\frac { 1 } { \sqrt { 15 } + \sqrt { 15 } } = \frac { 1 } { 2\sqrt { 15 } } = \frac { \sqrt { 15 } } { 2\sqrt { 15 } \cdot \sqrt { 15 } } = \frac { \sqrt { 15 } } { 30 }$$Therefore, the correct answer is $\frac { \sqrt { 15 } } { 30 }$.

Find lim⁡y→015+y−15y\lim _ { y \rightarrow 0 } \frac { \sqrt { 15 + y } - \sqrt { 15 } } { y }limy→0​y15+y​−15​​

Find $\lim _ { y \rightarrow 0 } \frac { \sqrt { 15 + y } - \sqrt { 15 } } { y }$